<p>4c) g’(x) = 2 + f(x)
g"(x) = f’(x)</p>
<p>you want f’(x) = 0, or undefined
no horizontal tangents, f’(x) != 0
verticle tangent: x = -3, cusp: x = 0 are undefined for f’(x)
concavity changes at x = -3 (down quarter-circle to up) but not at x=0 (both sides are down).</p>
<p>Inflection: x = -3, only.</p>
<p>Concavity has to change for inflection (unlike critical points), this only occurs at x = -3</p>
<p>A parabola can go from increasing to decreasing and would still always be concave down (it would have the max at the vertex though, what your teacher described is how we define max/min)</p>
<p>Then again, lines don’t have concavity, so . . . maybe it is inflection. However, the graph looks concave down. Hmmm, I now doubt my answer :/</p>
<p>4b was 1.5 because g(x) = integral of f(x) + 2. since f(x) stops being positive at x = 1.5, it stops increasing there. g(x) has a maximum at that point.</p>
<p>? g(x) =integral of f(x) +2x and g’(x) = 2 + f(x)</p>
<p>The inflection point was at X=0, everyone. In this case, you want where g’(x) changes from increasing to decreasing. -3 doesn’t work because although it may be a point of inflection for f(x), that’s not what the question asked: it asked for where g(x) had a point of inflection.</p>
<p>Yeah, that’s what I meant. If you had an integral at x = 2.5, it would become negative.</p>
<p>@PioneerJones. Ahh I see. That cleared some stuff up. The difference between f(x) and g(x)… that makes sense.</p>
<p>Do you think the Collegeboard would be brutal enough to take a point off if I also considered -3… although I did put x=0 also as an answer?..</p>
<p>gator, you’re wrong</p>
<p>the line segment equation is equal to -2x + 3 right? you can even use your calculator to verify that 2.5 is infact the answer</p>
<p>type this into your ti89</p>
<p>2 ( 1.5 ) + integral( 2x + 3 , x , 0 , 1.5) = 5.25</p>
<p>2 ( 2.5 ) + integral ( 2x + 3 , x , 0 , 2.5 ) = 6.25</p>
<p>now tell me which one is bigger >.<</p>
<p>For the region R problem, #3, it says to find the perimeter and volume in terms of k. Does that mean we should be using k’s in the expressions or x’s? </p>
<p>I used k’s without thinking and a classmate was saying that he used x. I thought he was right, but when I went back just now, it says to do it in terms of k. What did you guys do?</p>
<p>Well, when it specifically says “in terms of k” I would suppose you’d do the problem in terms of K…</p>
<p>I did what you did Kevinnnnni.</p>
<p>probably should have used k to be safe</p>
<p>i got 1 + k + e^2k + integral of sqroot (1 + 4e^4k)</p>
<p>for the volume i got e^4k*pi/4 - pi/4</p>
<p>for dv/dt i got (pi * e^2) / 3</p>
<p>I used X’s in the integral, but my bounds were 0 and k. Technically, that’s correct.</p>
<p>Ok, good. Thanks! Is anyone planning on making a solution guide?</p>
<p>What were your answers for #5?</p>
<p>theyellowboss, its the integral from 0 to k of sqroot (1+4e^4k)</p>
<p>yeah forgot to type the limits of int, it was 0 to k yup</p>
<p>Preliminary answer guide, feel free to contribute.</p>
<p>3a)1+k+e^2k+Int(sqt(1+4e^4x)dx,0,k)
3b) I’ll have to rework this one
3c) Same here</p>
<p>4a)
4b)
4c) X=0
4d) -2/7; function must differentiable throughout the interval for the mean value theorem to apply, but f is not differentiable at 0</p>
<p>5a) 1411
5b) (1/625)*(W-300), underestimate
5c) 1100e^t/25+300</p>
<p>6a)
6b)
6c) -121
6d)</p>
<p>6a)
Sin(x) = x - x^³/3 + x^5/120 - x^7/5040
Sin(x²) = x² - x^6/3 + x^10/120 - x^14/5040</p>
<p>6b)
Cos(x) = 1 - x²/2 + x^4/24 - x^6/720
Sin(x²) + cos(x) = 1 + x²/2 + x^4/24 - 121x^6/720</p>
<p>6c) -121</p>
<p>6d) Max value of the 5th derivative from [0,1/4] was 30. Therefore, by using the Lagrange error: </p>
<p>30*(1/4)^5
------------ = 1/4096
(5)!</p>
<p>Therefore the maximum error is 1/4096, which is definitely less than 1/3000.
Not 100% sure, please double check me on this.</p>
<p>4a </p>
<p>g(3) = -6 - 9pi/4
g’(x) = f(x) + 2
g(-3) = 2</p>
<p>4b) absolute maximum = 2.5, lots of people thought it was 1.5 but you can easily confirm on the calculator that its 2.5 (work is shown on the page before this one and explanation on this page)</p>
<p>^ I got all the same answers for number 6 (except I left some of my series with factorials - shouldn’t matter though) So those are confirmed.</p>
<p>3a)1+k+e^2k+Int(sqt(1+4e^4x)dx,0,k)
3b) I’ll have to rework this one
3c) Same here</p>
<p>4a) -33pi/4??I’m not positive (idk if this is what I put on the test), g’(x) = 2+f(x), g’(-3) = 2+0 = 2
4b) I recall it was x=2.5 - the justification was a nightmare (I didn’t know how to explain every point I tested and the line formula and calculations took up a ridiculous amount of space)
4c) X=0; g’(x) is undefined here and f changes concavity. We can’t tell anything at x=-4 because the interval ends there, it could be a horizontal line for all we know - no change in concavity.
4d) -2/7; function must differentiable throughout the interval for the mean value theorem to apply, but f is not differentiable at 0</p>
<p>5a) 1411
5b) (1/625)*(W-300), underestimate
5c) 1100e^t/25+300</p>
<p>6a) Sin(x) = x - x^3/3 + x^5/120 - x^7/5040
Sin(x²) = x² - x^6/3 + x^10/120 - x^14/5040</p>
<p>6b)Cos(x) = 1 - x²/2 + x^4/24 - x^6/720
Sin(x²) + cos(x) = 1 + x²/2 + x^4/24 - 121x^6/720</p>
<p>6c) -121
6d)
Max value of the 5th derivative from [0,1/4] was 30. Therefore, by using the Lagrange error: </p>
<p>30*(1/4)^5
------------ = Something like 1/4096?
(5)!</p>
<p>Therefore the maximum error is 1/4096, which is definitely less than 1/3000.</p>
<p>^ those were copied from Weihao</p>
<p>Also, did we need to show our calculation for maximum value for 6d?</p>
<p>Welhao, I agree on all answers. theyellowboss, those were the answers I got. And I agree, it was 2.5</p>
<p>g’(x)=2+f(x)
0=2+f(x)
f(x)=-2
x=2.5</p>